Magnetic flux

A magnetic field has a direction, so the surface you push it through must have one too, or "how much passes" is not a question. Give every small patch an area vector dA: magnitude the patch's area in m², direction along the patch's normal. Only the component of B along that normal counts, so the flux through the patch is B⋅dA = B cos θ dA, and field lines running parallel to the surface slide along it contributing nothing. This is why a 5.0 cm × 5.0 cm card held edge-on in the 0.60 T gap of an electromagnet catches zero flux while the field at the card is as strong as anywhere in the machine. Turn the card face-on and the same 0.60 T gives Φ = 0.60 × 0.0025 = 1.5 × 10⁻³ Wb. The field did not change; the surface's direction did.

When B is uniform over a flat loop, B and n̂ are the same at every point of it, so B cos θ comes out of the integral and what remains is the area itself: Φ_B = B A cos θ. Read θ carefully — it runs from the field to the normal. A 0.040 m² loop in a 0.30 T field holds 12.0 mWb face-on, 10.4 mWb at θ = 30°, 6.0 mWb at θ = 60°, and exactly zero at θ = 90°, where the plane of the loop contains the field. Two readings of the same product are worth carrying forward. B cos θ is the normal component of the field acting over the full area; A cos θ is the full field acting over the projected area, the loop's shadow cast along B. They give the same number, and the shadow picture is the one that survives when the surface is not flat.

Nothing in the definition demands a uniform field; it demands that you find a way of slicing the surface so B is constant across each slice. The standard case is a rectangular loop lying in the plane of a long straight wire. The wire's field is B = μ₀I/(2πr), identical at every point the same distance r away, so slice the loop into strips parallel to the wire: a strip at distance r, of length L and thickness dr, has dA = L dr with B constant across it and n̂ parallel to B. Then Φ = ∫ (μ₀I/2πr) L dr from a to a + w = (μ₀ I L/2π) ln((a + w)/a). For I = 8.0 A, L = 0.30 m, a = 10 mm and w = 40 mm that is 2 × 10⁻⁷ × 8.0 × 0.30 × ln 5 = 7.7 × 10⁻⁷ Wb. The logarithm is where the nonuniformity lives, and it is worth reading before you trust any estimate: widening the loop from 40 mm to 80 mm replaces ln 5 by ln 9 and raises the flux by only 37%, because every strip you add sits further out in a weaker field.

The normal has two directions and the physics will not choose for you, so you choose, and then never change your mind mid-problem. The convention that keeps the rest of this section consistent is the right-hand rule applied to the rim: curl the fingers of your right hand the way you have decided to travel round the loop, and your thumb points along n̂. Flip the circulation and n̂ flips, so Φ and dΦ/dt both change sign — and so does the sense in which a positive current counts. Those two flips cancel, leaving the real current flowing the same way round the real wire. A negative flux therefore carries no absolute meaning: it records that the field crosses the surface against n̂, entering through the very face your thumb points out of and leaving through the face behind it. What matters is that one sense of going round the loop is fixed for the whole problem, because the next topic differentiates Φ and the sign of that answer is only as trustworthy as this choice.

Nothing has ever been found for magnetic field lines to start or end on — no magnetic charge — so the outward flux through any closed surface is exactly zero: ∮ B⋅dA = 0. Whatever enters must leave. The practical consequence is one you will use constantly. Take a wire loop and stretch two different surfaces across it, a flat disc and a bulging dome. Together they bound a closed volume, so the flux out through one equals the flux in through the other, and once both are oriented consistently with the rim they carry exactly the same Φ. Flux belongs to the boundary. So a circular rim of radius 4.0 cm lying square to a uniform 0.60 T field passes Φ = 0.60 × π × 0.040² = 3.0 × 10⁻³ Wb through the flat disc, and a hemispherical cap on that same rim passes the same 3.0 mWb, with no integration over the curved surface needed.

Because Φ = B A cos θ is a product, it changes when any factor changes, and in a real device usually more than one does at once. Differentiating gives dΦ/dt = (dB/dt) A cos θ + B (dA/dt) cos θ − B A sin θ (dθ/dt): a field that grows, an area that grows, an orientation that turns. Keep the three apart when you set a problem up, and name which one the apparatus actually varies. A square loop of side 0.15 m sitting square to a field ramping at 0.80 T s⁻¹ gains flux at 0.0225 m² × 0.80 T s⁻¹ = 1.8 × 10⁻² Wb s⁻¹. A loop of that same 0.15 m width sliding out of a 0.40 T region at 0.20 m s⁻¹ loses area at 0.030 m² s⁻¹ and so loses flux at 1.2 × 10⁻² Wb s⁻¹. One weber per second is one volt, which is the whole reason flux is the quantity worth tracking — but turning dΦ/dt into an emf is the next topic's work.